3.1289 \(\int \frac {(b d+2 c d x)^{3/2}}{a+b x+c x^2} \, dx\)

Optimal. Leaf size=117 \[ -2 d^{3/2} \sqrt [4]{b^2-4 a c} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-2 d^{3/2} \sqrt [4]{b^2-4 a c} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )+4 d \sqrt {b d+2 c d x} \]

[Out]

-2*(-4*a*c+b^2)^(1/4)*d^(3/2)*arctan((d*(2*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))-2*(-4*a*c+b^2)^(1/4)*d^(3
/2)*arctanh((d*(2*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))+4*d*(2*c*d*x+b*d)^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {692, 694, 329, 212, 206, 203} \[ -2 d^{3/2} \sqrt [4]{b^2-4 a c} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-2 d^{3/2} \sqrt [4]{b^2-4 a c} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )+4 d \sqrt {b d+2 c d x} \]

Antiderivative was successfully verified.

[In]

Int[(b*d + 2*c*d*x)^(3/2)/(a + b*x + c*x^2),x]

[Out]

4*d*Sqrt[b*d + 2*c*d*x] - 2*(b^2 - 4*a*c)^(1/4)*d^(3/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d
])] - 2*(b^2 - 4*a*c)^(1/4)*d^(3/2)*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 692

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(2*d*(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1))/(b*(m + 2*p + 1)), x] + Dist[(d^2*(m - 1)*(b^2 - 4*a*c))/(b^2*(m + 2*p + 1)), In
t[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[
2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] &
& RationalQ[p]) || OddQ[m])

Rule 694

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[x^m*(
a - b^2/(4*c) + (c*x^2)/e^2)^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 0]

Rubi steps

\begin {align*} \int \frac {(b d+2 c d x)^{3/2}}{a+b x+c x^2} \, dx &=4 d \sqrt {b d+2 c d x}+\left (\left (b^2-4 a c\right ) d^2\right ) \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )} \, dx\\ &=4 d \sqrt {b d+2 c d x}+\frac {\left (\left (b^2-4 a c\right ) d\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (a-\frac {b^2}{4 c}+\frac {x^2}{4 c d^2}\right )} \, dx,x,b d+2 c d x\right )}{2 c}\\ &=4 d \sqrt {b d+2 c d x}+\frac {\left (\left (b^2-4 a c\right ) d\right ) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b^2}{4 c}+\frac {x^4}{4 c d^2}} \, dx,x,\sqrt {d (b+2 c x)}\right )}{c}\\ &=4 d \sqrt {b d+2 c d x}-\left (2 \sqrt {b^2-4 a c} d^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b^2-4 a c} d-x^2} \, dx,x,\sqrt {d (b+2 c x)}\right )-\left (2 \sqrt {b^2-4 a c} d^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b^2-4 a c} d+x^2} \, dx,x,\sqrt {d (b+2 c x)}\right )\\ &=4 d \sqrt {b d+2 c d x}-2 \sqrt [4]{b^2-4 a c} d^{3/2} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )-2 \sqrt [4]{b^2-4 a c} d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 113, normalized size = 0.97 \[ \frac {2 (d (b+2 c x))^{3/2} \left (-\sqrt [4]{b^2-4 a c} \tan ^{-1}\left (\frac {\sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )-\sqrt [4]{b^2-4 a c} \tanh ^{-1}\left (\frac {\sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )+2 \sqrt {b+2 c x}\right )}{(b+2 c x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*d + 2*c*d*x)^(3/2)/(a + b*x + c*x^2),x]

[Out]

(2*(d*(b + 2*c*x))^(3/2)*(2*Sqrt[b + 2*c*x] - (b^2 - 4*a*c)^(1/4)*ArcTan[Sqrt[b + 2*c*x]/(b^2 - 4*a*c)^(1/4)]
- (b^2 - 4*a*c)^(1/4)*ArcTanh[Sqrt[b + 2*c*x]/(b^2 - 4*a*c)^(1/4)]))/(b + 2*c*x)^(3/2)

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fricas [B]  time = 1.14, size = 210, normalized size = 1.79 \[ 4 \, \sqrt {2 \, c d x + b d} d - 4 \, \left ({\left (b^{2} - 4 \, a c\right )} d^{6}\right )^{\frac {1}{4}} \arctan \left (\frac {\left ({\left (b^{2} - 4 \, a c\right )} d^{6}\right )^{\frac {3}{4}} \sqrt {2 \, c d x + b d} d - \left ({\left (b^{2} - 4 \, a c\right )} d^{6}\right )^{\frac {3}{4}} \sqrt {2 \, c d^{3} x + b d^{3} + \sqrt {{\left (b^{2} - 4 \, a c\right )} d^{6}}}}{{\left (b^{2} - 4 \, a c\right )} d^{6}}\right ) - \left ({\left (b^{2} - 4 \, a c\right )} d^{6}\right )^{\frac {1}{4}} \log \left (\sqrt {2 \, c d x + b d} d + \left ({\left (b^{2} - 4 \, a c\right )} d^{6}\right )^{\frac {1}{4}}\right ) + \left ({\left (b^{2} - 4 \, a c\right )} d^{6}\right )^{\frac {1}{4}} \log \left (\sqrt {2 \, c d x + b d} d - \left ({\left (b^{2} - 4 \, a c\right )} d^{6}\right )^{\frac {1}{4}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

4*sqrt(2*c*d*x + b*d)*d - 4*((b^2 - 4*a*c)*d^6)^(1/4)*arctan((((b^2 - 4*a*c)*d^6)^(3/4)*sqrt(2*c*d*x + b*d)*d
- ((b^2 - 4*a*c)*d^6)^(3/4)*sqrt(2*c*d^3*x + b*d^3 + sqrt((b^2 - 4*a*c)*d^6)))/((b^2 - 4*a*c)*d^6)) - ((b^2 -
4*a*c)*d^6)^(1/4)*log(sqrt(2*c*d*x + b*d)*d + ((b^2 - 4*a*c)*d^6)^(1/4)) + ((b^2 - 4*a*c)*d^6)^(1/4)*log(sqrt(
2*c*d*x + b*d)*d - ((b^2 - 4*a*c)*d^6)^(1/4))

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giac [B]  time = 0.21, size = 354, normalized size = 3.03 \[ -\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} d \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right ) - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} d \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right ) - \frac {1}{2} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} d \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right ) + \frac {1}{2} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} d \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right ) + 4 \, \sqrt {2 \, c d x + b d} d \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a),x, algorithm="giac")

[Out]

-sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*d*arctan(1/2*sqrt(2)*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4) + 2*sqrt(2*c*
d*x + b*d))/(-b^2*d^2 + 4*a*c*d^2)^(1/4)) - sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*d*arctan(-1/2*sqrt(2)*(sqrt(2
)*(-b^2*d^2 + 4*a*c*d^2)^(1/4) - 2*sqrt(2*c*d*x + b*d))/(-b^2*d^2 + 4*a*c*d^2)^(1/4)) - 1/2*sqrt(2)*(-b^2*d^2
+ 4*a*c*d^2)^(1/4)*d*log(2*c*d*x + b*d + sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*sqrt(2*c*d*x + b*d) + sqrt(-b^2*
d^2 + 4*a*c*d^2)) + 1/2*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*d*log(2*c*d*x + b*d - sqrt(2)*(-b^2*d^2 + 4*a*c*d
^2)^(1/4)*sqrt(2*c*d*x + b*d) + sqrt(-b^2*d^2 + 4*a*c*d^2)) + 4*sqrt(2*c*d*x + b*d)*d

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maple [B]  time = 0.05, size = 582, normalized size = 4.97 \[ \frac {4 \sqrt {2}\, a c \,d^{3} \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}}}-\frac {4 \sqrt {2}\, a c \,d^{3} \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}}}-\frac {2 \sqrt {2}\, a c \,d^{3} \ln \left (\frac {2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}}}-\frac {\sqrt {2}\, b^{2} d^{3} \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}}}+\frac {\sqrt {2}\, b^{2} d^{3} \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}}}+\frac {\sqrt {2}\, b^{2} d^{3} \ln \left (\frac {2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )}{2 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}}}+4 \sqrt {2 c d x +b d}\, d \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a),x)

[Out]

4*d*(2*c*d*x+b*d)^(1/2)+4*d^3/(4*a*c*d^2-b^2*d^2)^(3/4)*2^(1/2)*arctan(-2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c
*d*x+b*d)^(1/2)+1)*a*c-d^3/(4*a*c*d^2-b^2*d^2)^(3/4)*2^(1/2)*arctan(-2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*
x+b*d)^(1/2)+1)*b^2-2*d^3/(4*a*c*d^2-b^2*d^2)^(3/4)*2^(1/2)*ln((2*c*d*x+b*d+(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x
+b*d)^(1/2)*2^(1/2)+(4*a*c*d^2-b^2*d^2)^(1/2))/(2*c*d*x+b*d-(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)*2^(1
/2)+(4*a*c*d^2-b^2*d^2)^(1/2)))*a*c+1/2*d^3/(4*a*c*d^2-b^2*d^2)^(3/4)*2^(1/2)*ln((2*c*d*x+b*d+(4*a*c*d^2-b^2*d
^2)^(1/4)*(2*c*d*x+b*d)^(1/2)*2^(1/2)+(4*a*c*d^2-b^2*d^2)^(1/2))/(2*c*d*x+b*d-(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d
*x+b*d)^(1/2)*2^(1/2)+(4*a*c*d^2-b^2*d^2)^(1/2)))*b^2-4*d^3/(4*a*c*d^2-b^2*d^2)^(3/4)*2^(1/2)*arctan(2^(1/2)/(
4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+1)*a*c+d^3/(4*a*c*d^2-b^2*d^2)^(3/4)*2^(1/2)*arctan(2^(1/2)/(4*a*
c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+1)*b^2

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive, negative or zero?

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mupad [B]  time = 0.59, size = 97, normalized size = 0.83 \[ 4\,d\,\sqrt {b\,d+2\,c\,d\,x}-2\,d^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )\,{\left (b^2-4\,a\,c\right )}^{1/4}-2\,d^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )\,{\left (b^2-4\,a\,c\right )}^{1/4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*d + 2*c*d*x)^(3/2)/(a + b*x + c*x^2),x)

[Out]

4*d*(b*d + 2*c*d*x)^(1/2) - 2*d^(3/2)*atan((b*d + 2*c*d*x)^(1/2)/(d^(1/2)*(b^2 - 4*a*c)^(1/4)))*(b^2 - 4*a*c)^
(1/4) - 2*d^(3/2)*atanh((b*d + 2*c*d*x)^(1/2)/(d^(1/2)*(b^2 - 4*a*c)^(1/4)))*(b^2 - 4*a*c)^(1/4)

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sympy [B]  time = 34.62, size = 352, normalized size = 3.01 \[ - 16 a c d^{3} \operatorname {RootSum} {\left (t^{4} \left (16384 a^{3} c^{3} d^{6} - 12288 a^{2} b^{2} c^{2} d^{6} + 3072 a b^{4} c d^{6} - 256 b^{6} d^{6}\right ) + 1, \left (t \mapsto t \log {\left (16 t a c d^{2} - 4 t b^{2} d^{2} + \sqrt {b d + 2 c d x} \right )} \right )\right )} + 4 b^{2} d^{3} \operatorname {RootSum} {\left (t^{4} \left (16384 a^{3} c^{3} d^{6} - 12288 a^{2} b^{2} c^{2} d^{6} + 3072 a b^{4} c d^{6} - 256 b^{6} d^{6}\right ) + 1, \left (t \mapsto t \log {\left (16 t a c d^{2} - 4 t b^{2} d^{2} + \sqrt {b d + 2 c d x} \right )} \right )\right )} - 4 b d^{2} \operatorname {RootSum} {\left (t^{4} \left (1024 a c d^{2} - 256 b^{2} d^{2}\right ) + 1, \left (t \mapsto t \log {\left (256 t^{3} a c d^{2} - 64 t^{3} b^{2} d^{2} + \sqrt {b d + 2 c d x} \right )} \right )\right )} + 4 b d^{2} \operatorname {RootSum} {\left (t^{4} \left (1024 a c d^{2} - 256 b^{2} d^{2}\right ) + 1, \left (t \mapsto t \log {\left (256 t^{3} a c d^{2} - 64 t^{3} b^{2} d^{2} + \sqrt {b d + 2 c d x} \right )} \right )\right )} + 4 d \sqrt {b d + 2 c d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)**(3/2)/(c*x**2+b*x+a),x)

[Out]

-16*a*c*d**3*RootSum(_t**4*(16384*a**3*c**3*d**6 - 12288*a**2*b**2*c**2*d**6 + 3072*a*b**4*c*d**6 - 256*b**6*d
**6) + 1, Lambda(_t, _t*log(16*_t*a*c*d**2 - 4*_t*b**2*d**2 + sqrt(b*d + 2*c*d*x)))) + 4*b**2*d**3*RootSum(_t*
*4*(16384*a**3*c**3*d**6 - 12288*a**2*b**2*c**2*d**6 + 3072*a*b**4*c*d**6 - 256*b**6*d**6) + 1, Lambda(_t, _t*
log(16*_t*a*c*d**2 - 4*_t*b**2*d**2 + sqrt(b*d + 2*c*d*x)))) - 4*b*d**2*RootSum(_t**4*(1024*a*c*d**2 - 256*b**
2*d**2) + 1, Lambda(_t, _t*log(256*_t**3*a*c*d**2 - 64*_t**3*b**2*d**2 + sqrt(b*d + 2*c*d*x)))) + 4*b*d**2*Roo
tSum(_t**4*(1024*a*c*d**2 - 256*b**2*d**2) + 1, Lambda(_t, _t*log(256*_t**3*a*c*d**2 - 64*_t**3*b**2*d**2 + sq
rt(b*d + 2*c*d*x)))) + 4*d*sqrt(b*d + 2*c*d*x)

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